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Q6(b)(ii):Hey folks,
please explain me this question! Q6 part b1 and and part c of MAY/JUNE2009, Q9 of MAY/JUNE 2009, and Q10 c2( my answer is coming as 10V but in M.S it is 9.0V)
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf
F = BIL
mg = BIL
Substitute values
(2.3 × 10^-3) (9.8) = B × (2.6) (4.4 × 10^-2)
B = 0.20 T
Q6(c):
Peak value, I0 = 2.3 × √2 = 3.25 A
The a.c current has one peak at positive value and the other peak at negative value so,
Variation at positive peak = 3.25 g
Variation at negative peak = - 3.25 g
The total variation is
= (3.25) - (-3.25)
= 6.5 g
Q10(c)(ii):
Vin = +0.4 V
Vout = Gain * Vin
Vout = 25 * (+0.4)
Vout = 10 V
But this output is more than the supply of the op-amp, therefore the amplifier saturates and the output is equal to the supply voltage!
So,
Vout = 9 V